A) 13 m
B) 14 m
C) 15 m
D) 12.8 m
Correct Answer: A
Solution :
Let AB and CD be the poles such that |
AB = 6 m, CD = 11 m and BD = 12 m |
Draw \[AE\bot CD.\]Then, AE = BD = 12 m |
\[CE=CD-DE=CD-AB=(11-6)m=5\,\,\text{m}\] |
From right \[\Delta AEC,\]we have |
\[A{{C}^{2}}=A{{E}^{2}}+C{{E}^{2}}={{(12)}^{2}}+{{5}^{2}}=(144+25)=169\] |
\[\therefore \] \[AC=\sqrt{169}=13\,\,\text{cm}\] |
Distance between their tops = 13 m |
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